Ph-CH=CH2 [HBr](PhCOO)2 Product Consider the above reaction Identify the correct statements. Choose the correct answer from the options given below :
- A & E Only
- A, B & D Only
- A, C & E Only
- C, D & E Only
Answer
(3) A, C & E Only
Solution
In the presence of benzoyl peroxide, HBr adds to styrene by a free radical mechanism (peroxide effect):
Initiation: (C6H5COO)2 2C6H5COO 2C6H5 + 2CO2
C6H5 + HBr C6H6 + Br benzene is formed as a byproduct (C correct) and the peroxide generates Br , not H (B incorrect).
Propagation: Br adds to the terminal carbon so that the more stable benzylic radical is formed (A correct):
C6H5-CH=CH2 + Br C6H5- CH-CH2Br
C6H5- CH-CH2Br + HBr C6H5-CH2-CH2Br + Br
1-Bromo-2-phenylethane is the major (anti-Markovnikov) product, not the minor one (D incorrect).
Without peroxide, HBr adds by electrophilic addition through the more stable benzylic carbocation C6H5- CH-CH3, giving 1-bromo-1-phenylethane (E correct).
A, C and E are correct.
Answer: (3)